# why can we expand an expandable function for infinite?

I know that by letting $x=\frac1u$ and find the expanded form of the new function centered at $0$ we can find the expanded form of a function at infinity,

but when we calculating the taylor polynomial we are centered around a point no? infinity is not a point, it is a limit. so why letting infinity to be a point works? when we are setting $\lim_{x\to\infty}x=\lim_{u\to0^+}\frac1u$ we still have a limit and not a point, if i regard this as a point then i can say $\lim_{x\to\infty}x=\lim_{v\to0}\frac1{v^2}$.

using few examples i saw that the final result will stay the same regardless of the limit, but i still can't understand why: why can we use limit as a point and why the speed that the limit get closer to the point doesn't matter?

This is what you define as expanding at infinity.

We say $f(z)$ is analytic at $z=\infty$, if $g(w) = f(1/w)$ is analytic at $w=0$.

The reason is, we want to study the behaviour of the function as $z$ gets larger. As you pointed out, notion of Power Series Expansion is defined around a point and it does not make much sense to take $\infty$ as a point and expand. But we still want to study the behaviour of the function as $z$ gets larger. Then observe that it is equivalent to study $g(w)$ around $0$, since $1/w$ 'gets larger' as $w\to 0$, and define it as above.

In other words, what you wrote as a 'procedure'

I know that by letting $x=\frac{1}{u}$ and find the expanded form of the new function centered at $0$ we can find the expanded form of a function at infinity,

Is actually the definition.

If we agree on that expanding a function around $0$ has no relation with the (speed)? of some limit, it is also irrelevant to expending at $\infty$ by above definition.